Question

There is a pendulum executing simple harmonic motion and its maximum kinetic energy is $K_1.$ If the length of the pendulum is doubled and it performs simple harmonic motion with the same angular amplitude as in the first case, its maximum kinetic energy is $K_2$ . Then :

• A. $K_2$
• B. $K_2=2K_1$
• C. $K_2=K_1$
• D. $K_2=\frac{k_1}{4}$

Answer 1

Answer: (B)

Maximum kinetic energy $=\frac{1}{2}m{\omega }^2{A}^2$
$\omega =\sqrt{\frac{g}{L}}$
$A=L\theta$
$KE=\frac{1}{2}m\frac{g}{L}\times L^2{\theta }^2,KE=\frac{1}{2}mgL{\theta }^2$
$K_1=\frac{1}{2}mgL{\theta }^2$
If Length Is Double
$K_2=\frac{1}{2}mg(2L){\left(\theta \right)}^2$ [Here we are assuming angular amplitude is same]
$\frac{K_1}{K_2}=\frac{\frac{1}{2}mgl{\left(\theta \right)}^2}{\frac{1}{2}mg(2L){\left(\theta \right)}^2}=\frac{1}{2}$
$K_2=2K_1$